If the line $3x-2y+6=0$ meets X-axis and Y-axis respectively at A and B, then the equation of the circle with radius AB and centre at A is

The given line is $3x-2y+6=0$

This line meets the $X-$ axis at $A\left(-2,0\right)$ and the above line cuts the $Y-$ axis at $B\left(0,3\right)$

The radius of the required circle is 

   $\begin{array}{rcl}r& =& AB\\ & =& \sqrt{4+9}\\ & =& \sqrt{13}\end{array}$

The center of the circle is $A\left(-2,0\right)$

The equation of the circle having cetne at $A\left(-2,0\right)$ and radius $r=\sqrt{13}$ is 

${\left(x+2\right)}^2+y^2=13\Rightarrow \boxed{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{9}\mathbf{=}\mathbf{0}}$

(or)

$$\begin{gathered} \mathrm{meets} \mathrm{x}-\mathrm{ax}is put \mathrm{y}=0 then \\ \mathrm{A}=(-2, 0) \\ \mathrm{meets} \mathrm{y}-\mathrm{ax}is then \mathrm{x}=0 then \\ \mathrm{B}(0, 3) \\ \mathrm{r}=\mathrm{AB}=\sqrt{9+4}=\sqrt{13} \\ {so circle is (\mathrm{x}+2)}^2+{\mathrm{y}}^2=13 \\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-9=0 \end{gathered}$$